Let $X_1,...,X_n$ be an iid (independent and identically distributed) sample with mean $ \mu $ and variance $\sigma^2$.
We can use this conclusion ：$ (n-1)S^2 = \sum_{i=1}^n (Xi-\overline X)^2 = \sum_{i=1}^n (Xi-\mu ) ^2 - n(\mu-\overline X) ^2 $
Suppose that $\mathbb E(X_i-\mu)^4<\infty$, and use the Weak Law of Large Numbers to show that
$ S^2 \rightarrow \sigma^2$ (in probability) as $n \rightarrow \infty $
My questions are:
How do I go about showing that?
Why suppose $\mathbb E(X_i-\mu)^4<\infty$? (The second may be included in the first)
Hint: Since $\ \left(X_i-\mu\right)^2, i=1,2,\dots $ are iid random variables, the weak law of large numbers applies to them, provided they have finite variance. That is, under the stated condition, $\ \frac{1}{n}\sum_{i=1}^n (Xi-\mu )^2\ =\left(1-\frac{1}{n}\right)S^2+$ $\left(\mu-\overline X\right)^2\ $ converges in probability to $\ \mathbb{E}\left(Xi-\mu \right)^2=\sigma^2\ $, and you also know that $\ \left(\mu-\overline X\right)^2 $ converges to $\ 0\ $ in probability. So, what is the relation between the variance of $\ \left(X_i-\mu\right)^2\ $ and $\ \mathbb{E}\left(Xi-\mu \right)^4\ $?